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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2017 · Paper II

Q1(d). LDA at -78°C: kinetic enolate of 2-phenylcyclohexanone

  • 10 marks
  • Answered
Question

Give the structures of C and D and account for their formation.
[structure] → (LiN(iPr)₂, -78°C, THF) C → (MeI) D (major)

Answer in brief

C = kinetic lithium enolate formed by removal of the C-6 hydrogen; D = 2-methyl-6-phenylcyclohexan-1-one (major). The more substituted 2-methyl-2-phenyl ketone would need thermodynamic conditions.

Key points

  • C. The C-2 hydrogen is more acidic (benzylic) but is tertiary and shielded by the phenyl ring. The bulky (iPr)₂N⁻ removes a less hindered C-6 hydrogen through a cyclic six-membered transition state with Li bound to the…
  • At -78°C with no excess ketone there is no proton exchange, so C does not equilibrate to the more stable, conjugated (thermodynamic) enolate C1=C2–Ph.
  • D. The enolate α-carbon (C-6) attacks CH₃I by SN2 (C-alkylation): D = 2-methyl-6-phenylcyclohexan-1-one (the new methyl and the phenyl on the two different α-carbons; a cis/trans mixture that favours the diequatorial isomer on…
  • Under thermodynamic conditions (NaH or Et₃N/Me₃SiCl with heating, or excess ketone) the conjugated enolate forms instead and methylation gives 2-methyl-2-phenylcyclohexanone.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. 2-Phenylcyclohexanone has two sets of α-hydrogens: one tertiary, benzylic H on C-2 and two secondary H on C-6. LDA is a strong, bulky, non-nucleophilic base; at -78°C in THF deprotonation is fast and irreversible, so the enolate formed fastest (kinetic enolate) is the one obtained.

Scheme: LDA removes the more accessible C-6 hydrogen; the enolate carbon attacks MeI.

  1. C. The C-2 hydrogen is more acidic (benzylic) but is tertiary and shielded by the phenyl ring. The bulky (iPr)₂N⁻ removes a less hindered C-6 hydrogen through a cyclic six-membered transition state with Li bound to the carbonyl oxygen. C = lithium 6-phenylcyclohex-1-en-1-olate (the less substituted enolate).
  2. At -78°C with no excess ketone there is no proton exchange, so C does not equilibrate to the more stable, conjugated (thermodynamic) enolate C1=C2–Ph.
  3. D. The enolate α-carbon (C-6) attacks CH₃I by SN2 (C-alkylation): D = 2-methyl-6-phenylcyclohexan-1-one (the new methyl and the phenyl on the two different α-carbons; a cis/trans mixture that favours the diequatorial isomer on equilibration).
  4. Under thermodynamic conditions (NaH or Et₃N/Me₃SiCl with heating, or excess ketone) the conjugated enolate forms instead and methylation gives 2-methyl-2-phenylcyclohexanone.

Conclusion. C = kinetic lithium enolate formed by removal of the C-6 hydrogen; D = 2-methyl-6-phenylcyclohexan-1-one (major). The more substituted 2-methyl-2-phenyl ketone would need thermodynamic conditions.

Application. Kinetic (LDA, -78°C) versus thermodynamic enolate control decides the site of alkylation and aldol reactions in steroid and terpene syntheses.

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Model answer to Q1(d), 2017 Paper II: solution page 4
Solution page 4 of 23